Verifing that $f$ is integrable $f:[0,1]\times[0,1]\to \Bbb R$ be given by
$$f(x,y) =
\begin{cases}
xy^2,  & \ y\lt x^2 \\
x+2y, & \ y\ge x^2
\end{cases}$$
I need to show that $f$ is integrable. My idea is that to show $Dincont(f)\subseteq S={(x,y): y=x^2}$. If $S$ is measure $0$, $\implies$ discontinuities of $f$ is also measure $0$. The it would mean that $f$ is integrable. Can someone please give me a hint about how to show that $S$ is measure $0$?
Thanks in advance!
 A: The set $y=x^2$ in $[0,1] \times [0,1]$ can be shown to have measure zero if it can be contained in sets of arbitrarily small measure. So divide up each axis into $n$ parts, getting a grid with $n^2$ boxes each of area $1/n^2.$ The part needed to cover the graph of $y=x^2$ is bounded by a constant times $n$, say $kn$, so the area needed to cover the parabola is at most $kn/n^2=k/n,$ which can be made arbitrarily small.
That $k=3$ here works can be seen by noting that, above each interval $I_j=[(j-1)/n,j/n]$ for $1\le j \le n$ the graph of $y=x^2$ can cut through at most three of the vertically stacked squares of sidelength $1/n$ which lie above interval $I.$ This is dependent on the max slope of $y=x^2$ being $2$ in the interval $[0,1].$
Note: As outlined in a comment, the "max vertical boxes $3$" statement can be proved using the mean value theorem. For a specific case where $3$ vertical boxes are actually needed, consider $n=10$ and the interval $[.8,.9]$ with the squared endpoints $.64,\ .81.$ In this case the curve $y=x^2$ enters each of the three vertically aligned boxes in $[.8,.9] \times [.6,.9].$
